chapter 4: the mathematics of sharing 4.1 apportionment problems and apportionment methods

Chapter 4:The Mathematics of Sharing4.1 Apportionment Problems and Apportionment Methods

Excursions in Modern Mathematics, 7e: 1.1 - 2Copyright © 2010 Pearson Education, Inc.Copyright © 2014 Pearson Education. All rights reserved. 4.1-2

Without a doubt, the most important and heated debate at the Constitutional Convention concerned the makeup of the legislature. The small states wanted all states to have the same number of representatives; the larger states wanted some form of proportional representation. The Constitutional compromise, known asthe Connecticut Plan, was a Senate, in which every state has two senators, and a House of Representatives, in which each state has a number of representatives that is a function of its population.

Apportionment


While the Constitution makes clear that seats in the House of Representatives are to be allocated to the states based on their populations (“According to their respective Numbers”), it does not prescribea specific method for the calculations. Undoubtedly, the Founding Fathers felt that this was a relatively minor detail–a matter of simple arithmetic that could be easily figured out and agreed upon by reasonable people.

Apportionment


Certainly it was not the kind of thing to clutter a Constitution with or spend time arguing over in the heat of the summer. What the Founding Fathers did not realize is that Article I, Section 2, set the Constitution of the United States into a collision course with a mathematical iceberg known today(but certainly not then) as the apportionment problem.

Apportionment


Obviously, the word apportion is the key word in this chapter. There are two critical elements in thedictionary definition of the word: (1) We are dividing and assigning things, and (2) we are doing this on a proportional basis and in a planned, organized fashion.

We will start this section with a pair of examples that illustrate the nature of the problem we aredealing with. These examples will raise several questions but not give any answers. Most of the answers will come later.

Apportionment Problems


Mom has a total of 50 identical pieces of candy (let’s saycaramels), which she is planning to divide among her five children (this is the division part). Like any good mom, she is intent on doing this fairly. Of course, the easiest thing to do would be to give each child 10 caramels–by most standards, that would be fair. Mom, however, is thinking of the long-term picture–she wants to teach her children about the value of work and aboutthe relationship between work and reward.

Example 4.1Kitchen Capitalism


This leads her to the following idea: She announcesto the kids that the candy is going to be divided atthe end of the week in proportion to the amount of time each of them spends helping with the weekly kitchen chores–if you worked twice as long as your brother you get twice as much candy, and so on (this is the due and proper proportion part). Unwittingly, mom has turned this division problem into an apportionment problem.



At the end of the week, the numbers are in. Table 4-1 shows the amount of work done by each child during the week. (Yes, mom did keep up-to-the-minute records!)



According to the ground rules, Alan, who worked 150 out of a total of 900 minutes, should get 8 1/3 pieces.

Here comes the problem: Since the pieces of candy are indivisible, it is impossible for Alan to get his pieces–he can get 8 pieces (and get shorted) or he can get 9 pieces (and someone else will get shorted).



A similar problem occurs with each of the other children. Betty’s exact fair share should be 4 1/3 pieces; Connie’s should be 9 11/18 pieces; Doug’s, 11 1/3 pieces; and Ellie’s, 16 7/18 pieces.

Because none of these shares can be realized, an absolutely fair apportionment of the candy is going to be impossible.

What should mom do?



Example 4.1 shows all the elements of an apportionment problem–there are objects to be divided (the pieces of candy), and there is a proportionality criterion for the division (number of minutes worked during the week). We will say that the pieces of candy are apportioned to the kids, and we will describe the final solution as an apportionment (Alan’s apportionment is x pieces, Betty’s apportionment is y pieces, etc.).

Let’s now consider a seemingly different apportionment problem.

Apportionment Problem


It is the year 2525, and all the planets in the Utopia galaxy have finally signed a peace treaty. Five of the planets (Alanos, Betta, Conii, Dugos, and Ellisium) decide to join forces and form an Intergalactic Federation. The Federation will be ruled by an Intergalactic Congress consisting of 50 “elders,” and the 50 seats in the Intergalactic Congress are to beapportioned among the planets according to theirrespective populations.

Example 4.2 The Intergalactic Congress of Utopia


The population data for each of the planets (in billions) are shown in Table 4-2. Based on these population figures, what is the correct apportionment of seats to each planet?

Example 4.2 The Intergalactic Congress of Utopia

We’ll get back to this later . . .


Example 4.2 is another example of an apportionment problem–here the objects being divided are the 50 seats in the Intergalactic Congress, and the proportionality criterion is population. But there is more to it than that. The observant reader may have noticed that the numbers in Examples 4.1 and 4.2 are identical– it is only the framing of the problem that has changed. Mathematically speaking, Examples 4.1 and 4.2 are one and the same apportionment problem–solve either one and you have solved both!

Apportionment Problem


Parador is a small republic located in Central America and consists of six states: Azucar, Bahia, Cafe, Diamante, Esmeralda, and Felicidad (A, B, C, D, E, and F for short). There are 250 seats in the Congress, which, according to the laws of Parador,are to be apportioned among the six states in proportion to their respective populations. What is the “correct” apportionment? Table 4-3 shows the population figures for the six states according to the most recent census.

Example 4.3 The Congress of Parador


The first step we will take to tackle this apportionment problem is to compute the population to seats ratio, calledthe standard divisor (SD). In the case of Parador, the standard divisor is 12,500,000/250 = 50,000. The standard divisor tells us that in Parador, each seat in the Congress corresponds to 50,000 people.



We can now use this yardstick to find the number of seats that each state should get by the proportionality criterion– all we have to do is divide the state’s population by 50,000. For example, take state A. If we divide the population of A by the standard divisor, we get 1,646,000/50,000 = 32.92. This number is called the standard quota (sometimes also known as the exactor fair quota) of state A.



If seats in the Congress could be apportioned in fractional parts, then the fair and exact apportionment to A would be 32.92 seats, but, of course, this is impossible–seats in theCongress are indivisible objects! Thus, the standard divisor of 32.92 is a standard that we cannot meet, although in this case we might be able to come close to it: The next best solution might be to apportion 33 seats to state A. Hold that thought!



Using the standard divisor SD = 50,000, we can quickly find the standard quotas of each of the other states.These are shown in Table 4-4 (rounded to two decimal places). Notice that the sum of the standard quotas equals 250, the number of seats being apportioned.



Once we have computed the standard quotas, we getto the heart of the apportionment problem: How should we round these quotas into whole numbers? At first glance, this seems like a dumb question. After all, we all learned in school how to round decimals to whole numbers–round down if the fractional part is less than 0.5, round up otherwise. This kind of rounding is called rounding to the nearest integer, or simply conventional rounding.



Unfortunately, conventional rounding will not work in this example, and Table 4-5 shows why not–we would end up giving out 251 seats in the Congress, and there are only 250 seats to give out! (Add an extra seat, you say? Sorry, this is not a banquet. The number of seats is fixed–we don’t have the luxury of adding or subtracting seats!)



So now that we know that conventional rounding of the standard quotas is not going to give us the solution to the apportionment of Parador’s Congress, what do we try next? For now, we leave this example–a simple apportionment problem looking for a simple solution–with no answers (but, we hope, with some gained insight!). Our search for a good solution to this problem will be the theme of our journey through the rest of this chapter.



Apportionment problems can arise in a variety of real-life applications–dividing candy among children (see Example 4.1), assigning nurses to shifts (see Exercise 4.4), assigning buses to routes (see Exercise 4.3), and so on. But the gold standard for apportionment applications is the allocation of seats in a legislature, and thus it is standard practice to borrow the terminology of legislative apportionment and apply it to apportionment problems in general.

Apportionment: Basic Concepts and Terminology


The basic elements of every apportionment problem are as follows:

The “states”This is the term we will use to describe the parties having a stake in the apportionment. Unless they have specific names (Azucar, Bahia, etc.), we will use A1, A2,…, AN, to denote the states.



The “seats”This term describes the set of M identical, indivisible objects that are being divided among the N states. For convenience, we will assume that there are more seats than there are states, thus ensuring that every state can potentially get a seat. (This assumption does not imply that every state must get a seat!)



The “populations”This is a set of N positive numbers (for simplicity we will assume that they are whole numbers) that are used as the basis for the apportionment of the seats to the states. We will use p1, p2,…, pN, to denote the state’s respective populations and P to denote the total population P = p1 + p2 +…+ pN).



Two of the most important concepts of the chapter are the standard divisor and the standard quotas.

We can now formally define these concepts using our new terminology and notation.



The standard divisor(SD)This is the ratio of population to seats. It gives us a unit of measurement (SD people = 1 seat) for our apportionment calculations.



The standard quotasThe standard quota of a state is the exact fractional number of seats that the state would get if fractional seats were allowed. We will use the notation q1, q2,…,

qN to denote the standard quotas of the respective states. To find a state’s standard quota, we divide the state’s population by the standard divisor. (In general, the standard quotas can be expressed as fractions or decimals–round them to two or three decimal places.)



Upper and lower quotas

The lower quota is the standard quota rounded down and the upper quota is the standard quota rounded up. In the unlikely event that the standard quota is a whole number, the lower and upper quotas are the same. We will use L’s to denote lower quotas and U’s todenote upper quotas. For example, the standard quota q1 = 32.92 has lower quota L1 = 32 and upper

quota U1 = 33.



Our main goal in this chapter is to discover a “good” apportionment method–a reliable procedure that (1) will always produce a valid apportionment (exactly Mseats are apportioned) and (2) will always produce a“fair” apportionment. In this quest we will discuss several different methods and find out what is good and bad about each one.

Apportionment Methods


Chapter 4:The Mathematics of Sharing4.2 Hamilton’s Method


Hamilton’s method can be described quite briefly:

Every state gets at least its lower quota. As many states as possible get their upper quota, with the one with highest residue (i.e.,fractional part) having first priority, the one with second highest residue second priority, and so on.

A little more formally, it goes like this:

Hamilton’s Method


Step 1 Calculate each state’s standard quota.

Step 2 Give to each state its lower quota.

Step 3 Give the surplus seats (one at a time) to the states with the largest residues (fractional parts) until there are no more surplus seats.

HAMILTON’S METHOD


As promised, we are revisiting the Parador Congress apportionment problem. We are now going to find our first real solution to this problem–the solution given by Hamilton’s method (sometimes, for the sake of brevity, we will call it the Hamilton apportionment). Table 4-6 shows all the details and speaks for itself. (Reminder: The standard quotas in the second row of the table were computed in Example 4.3.)

Example 4.4Parador’s Congress(Hamilton’s Method)


Hamilton’s method (also known as Vinton’s method or the method of largest remainders) was used in the United States only between 1850 and 1900.

Example 4.4Parador’s Congress(Hamilton’s Method)


Hamilton’s method is still used today to apportion the legislatures of Costa Rica, Namibia, and Sweden.At first glance, Hamilton’s method appears to be quite fair. It could be reasonably argued that Hamilton’s method has a major flaw in the way it relies entirely on the size of the residues without consideration of what those residues represent as a percent of the state’s population. In so doing, Hamilton’s method creates a systematic bias in favor of larger states over smaller ones.

Hamilton’s Method


This is bad–a good apportionment method should be population neutral, meaning that it should not be biased in favor of large states over small ones or vice versa.

To be totally fair, Hamilton’s method has two important things going for it: (1) It is very easy to understand, and (2) it satisfies an extremely important requirement for fairness called the quota rule.

Hamilton’s Method


No state should be apportioned a number of seats smaller than its lower quota or larger than its upper quota. (When a state is apportioned a number smaller than its lower quota, we call ita lower-quota violation; when a state is apportioned a number larger than its upper quota, we call it an upper-quota violation.)

QUOTA RULE


An apportionment method that guarantees that everystate will be apportioned either its lower quota or its upper quota is said to satisfy the quota rule. It is not hard to see that Hamilton’s method satisfies the quota rule: Step 2 of Hamilton’s method hands out to each state its lower quota. Right off the bat this guarantees that there will be no lower-quota violations. In Step 3 some states get one extra seat, some get none; no state can get more than one. This guarantees that there will be no upper-quota violations.

Hamilton’s Method


Chapter 4:The Mathematics of Sharing4.3 Jefferson’s Method


Jefferson’s method is based on an approach very different from Hamilton’s method. It is in the handling of the surplus seats (Step 3) that Hamilton’s method runs into trouble. So here is an interesting idea: Let’s tweak things so that when the quotas are rounded down, there are no surplus seats! The answer is by changing the divisor, which then changes the quotas. The idea is that by using a smaller divisor, we make the quotas bigger.

Let’s work an example.

Jefferson’s Method


When we divide the populations by the standard divisor SD = 50,000, we get the standard quotas (third row of Table). When these quotas are rounded down, we end up with four surplus seats (fourth row). Now here are some new calculations: When we divide the populations by the slightly smaller divisor D = 49,500, we get a slightly bigger set of quotas (fifth row). When these modified quotas are rounded down (last row), the surplus seats are gone and we have a valid apportionment.

Example 4.8Parador’s Congress(Jefferson’s Method)


It’s nothing short of magical! And, of course, you are wondering, where did that 49,500 come from? Let’s call it a lucky guess for now.

Example 4.8Parador’s Congress(Jefferson’s Method)


Example 4.8 illustrates a key point: Apportionments don’t have to be based exclusively on the standard divisor.Jefferson’s method is but one of a group of apportionment methods based on the principle that the standard yardstick 1 seat = SD of people is not set in concrete and that, if necessary, we can change to a different yardstick: 1 seat = D people, where D is a suitably chosen number.

Apportionment and Standard Divisor


The number D is called a divisor (sometimes we use the term modified divisor), and apportionment methods that use modified divisors are called divisor methods. Differentdivisor methods are based on different philosophies of how the modified quotas should be rounded to whole numbers, but they all follow one script: When you are done rounding the modified quotas, all M seats have been apportioned (no more, and no less). To be able to do this, you just need to find a suitable divisor D.

Modified Divisor and Divisor Methods


Step 1 Find a “suitable” divisor D.

Step 2 Using D as the divisor, compute each state’s modified quota (modified quota = state population/D).

Step 3 Each state is apportioned its modified lower quota.

JEFFERSON’S METHOD


How does one find a suitable divisor D? There are different ways to do it, we will use a basic, blue-collar approach: educated trial and error. Our target is a set of modified lower quotas whose sum is M. For the sum ofthe modified lower quotas to equal M, we need to make the modified quotas somewhat bigger than thestandard quotas. This can only be accomplished by choosing a divisor somewhat smaller than SD. (As the divisor goes down, the quota goes up, and vice versa.)

Finding D


• Make an educated guess, choose a divisor D smaller than SD.

• If guess works, we’re done.

• If sum or lower quotas is less than M, choose even smaller value for D.

• If sum or lower quotas is more than M, choose a bigger value for D.

• Repeat this trial-and-error approach to find a divisor D that works.

Here’s a flow chart . . . .

Finding D


How does one find a suitable divisor D?

Finding D


The first guess should be a divisor somewhat smaller thanSD = 50,000. Start with D = 49,000. Using this divisor, we calculate the quotas, round them down, and add. We get a total of T = 252 seats. We overshot our target by two seats! Refine our guess by choosing a larger divisor D (the point is to make the quotas smaller). A reasonable next guess (halfway between 50,000 and 49,000) is 49,500. We go through the computation, and it works!

Example 4.8 Parador’s Congress(Jefferson’s Method)


By design Jefferson’s method is consistent in its approach to apportionment–the same formula applies to every state (find a suitable divisor, find the quota, round it down). By doing this, Jefferson’s method is able to avoid the major pitfalls of Hamilton’s method. There is, however, a serious problem with Jefferson’s method that can surface when we least expect it–it can produce upper-quotaviolations!–which tend to consistently favor the larger states.

Jefferson’s Method and Quota Rule


Chapter 4:The Mathematics of Sharing4.4 Adams’s and Webster’s Methods


Like Jefferson’s method, Adams’s method is a divisor

method, but instead of rounding the quotas down, it

rounds them up. For this to work the modified quotas

have to be made smaller, and this requires the use of a

divisor D larger than the standard divisor SD.

Adams’s Method




Step 3 Each state is apportioned its modified upper quota.

ADAMS’S METHOD


By now you know the background story by heart. The populations are shown once again in the second row of Table 4-15. The challenge, as is the case with any divisor method, is to find a suitable divisor D. We know that in Adams’s method, the modified divisor D will have to bebigger than the standard divisor of 50,000. We start withthe guess D = 50,500. The total is T = 251, one seat aboveour target of 250, so we need to make the quotas just a tad smaller.

Example 4.9Parador’s Congress(Adams’s Method)


Increase the divisor a little bit, try D = 50,700. This divisor works! The apportionment under Adams’s method is shown in the last row.

Example 4.9Parador’s Congress(Adams’s Method)


Example 4.9 highlights a serious weakness of this

method–it can produce lower-quota violations!

This is a different kind of violation, but just as serious

as the one in Example 4.8 –state B got 1.72 fewer

seats than what it rightfully deserves! We can

reasonably conclude that Adams’s method is no

better (or worse) than Jefferson’s method–just

different.

Adams’s Method and the Quota Rule


What is the obvious compromise between rounding all the quotas down (Jefferson’s method) and rounding all the quotas up (Adams’ method)? What about conventional rounding (Round the quotas down when the fractional part is less than 0.5 and up otherwise.)?

Now that we know that we can use modified divisors to manipulate the quotas, it is always possible to find a suitable divisor that will make conventional rounding work. This is the idea behind Webster’s method.

Webster’s Method




Step 3 Find the apportionment by rounding each modified quota the conventional way.

WEBSTER’S METHOD


Our first decision is to make a guess at the divisor D: Should it be more than the standard divisor (50,000), or should it be less? Use the standard quotas as a starting point. When we round off the standard quotas to the nearest integer, we get a total of 251 (row 4 of Table 4-16). This number is too high (just by one seat), which tells us that we should try a divisor Da tad larger than the standard divisor. We tryD = 50,100.

Example 4.10Parador’s Congress(Webster’s Method)


Row 5 shows the respective modified quotas, and the last row shows these quotas rounded to the nearest integer. Now we have a valid apportionment! The last row shows the final apportionment under Webster’s method.

Example 4.9Parador’s Congress(Adam’s Method)


A flowchart illustrating how to find a suitable divisor D

for Webster’s method using educated trial and error

is on the next slide. The most significant difference

when we use trial and error to implement Webster’s

method as opposed to Jefferson’s method is the

choice of the starting value of D. With Webster’s

method we always start with the standard divisor SD.

If we are lucky and SD happens to work, we are

done!

Webster’s Method


Webster’s Method


When the standard divisor works as a suitable

divisor for Webster’s method, every state gets an

apportionment that is within 0.5 of its standard

quota. This is as good an apportionment as one

can hope for. If the standard divisor doesn’t quite

do the job, there will be at least one state with an

apportionment that differs by more than 0.5 from its

standard quota.

Webster’s Method


In general, Webster’s method tends to produce

apportionments that don’t stray too far from the

standard quotas, although occasional violations of

the quota rule (both lower- and upper-quota

violations), but such violations are rare in real-life

apportionments. are possible. Webster’s method has

a lot going for it–it does not suffer from any

paradoxes, and it shows no bias between small and

large states.

Webster’s Method


Surprisingly, Webster’s method had a rather short

tenure in the U.S.House of Representatives. It was

used for the apportionment of 1842, then replaced

by Hamilton’s method, then reintroduced for the

apportionments of 1901, 1911, and 1931, and then

replaced again by the Huntington-Hill method, the

apportionment method we currently use.

Webster’s Method


Chapter 4:The Mathematics of Sharing4.6 The Quota Rule and Apportionment Paradoxes


The most serious (in fact, the fatal) flaw of Hamilton’smethod is commonly known as the Alabama paradox. In essence, the Alabama paradox occurs when an increase in the total number of seats being apportioned, in and of itself, forces a state to lose one of its seats. The best way to understand what this means is to look carefully at the following example.

Alabama Paradox - Hamilton’s Method


The small country of Calavos consists of three states: Bama, Tecos, and Ilnos. With a total population of 20,000 and 200 seats in the House of Representatives, the apportionment of the 200 seats under Hamilton’s method is shown in Table 4-7.

Example 4.5 More Seats Means Fewer Seats


Now imagine that overnight the number of seats is increased to 201, but nothing else changes. Since there is one more seat to give out, the apportionment has to be recomputed.



Here’s the new apportionment (Hamilton’s method) for a House with 201 seats. (Notice that for M = 200, the SD is 100; for M = 201, the SD drops to 99.5.)



The shocking part of this story is the fate of Bama, the “little guy.” When the House of Representatives had 200 seats, Bama got 10 seats, but when the number of seats to be divided increased to 201, Bama’s apportionment went down to 9 seats. How did this paradox occur? Notice the effect of the increase in M on the size of the residues: In a House with 200 seats, Bama is at the head of the priority line for surplus seats, but when the number of seats goes up to 201, Bama gets shuffled to the back of the line.



Example 4.5 illustrates the quirk of arithmetic behind

the Alabama paradox: When we increase the

number of seats to be apportioned, each state’s

standard quota goes up, but not by the same

amount. As the residues change, some states can

move ahead of others in the priority order for the

surplus seats. This can result in some state or states

losing seats they already had.

Arithmetic of the Alabama Paradox


Sometime in the early 1900s it was discovered that under Hamilton’s method a state could potentially lose some seats because its population got too big! This phenomenon is known as the population paradox. To be more precise, the population paradox occurs when state A loses a seat to state B even though the population of A grew at a higher rate than the population of B. Too weird to be true, you say? Check out the next example.

The Population Paradox


In the year 2525 the five planets in the Utopia galaxyfinally signed a peace treaty and agreed to form an Intergalactic Federation governed by an Intergalactic Congress. This is the story of the two apportionments that broke up the Federation.

Example 4.6 A Tale of Two Planets


Part I. The Apportionment of 2525.

The first Intergalactic Congress was apportioned using Hamilton’s method, based on the population figures (in billions) shown in the second column of Table 4-10. (This is the apportionment problem discussed in Example 4.2.) There were 50 seats apportioned.



Part I. The Apportionment of 2525.



Let’s go over the calculations. Since the total population of the galaxy is 900 billion, the standard divisor is SD = 900/50 = 18 billion. Dividing the planet populations by this standard divisor gives the standard quotas shown in the third column of Table 4-10. After the lower quotas are handed out (column 4), there are two surplus seats. The first surplus seat goes to Conii and the other one to Ellisium. The last column shows the apportionments. (Keep an eye on the apportionments of Betta and Ellisium–they are central to how this story unfolds.)



Part II. The Apportionment of 2535.

After 10 years of peace, all was well in the Intergalactic Federation. The Intergalactic Census of 2535 showed only a few changes in the planets’ populations–an 8 billion increase in the population ofConii, and a 1 billion increase in the population of Ellisium. The populations of the other planets remained unchanged from 2525. Nonetheless, a newapportionment was required.



Part II. The Apportionment of 2535.Here are the details of the 2535 apportionment under Hamilton’s method.



Notice that the total population increased to 909 billion, so the standard divisor for this apportionment was SD = 909/50 = 18.18.

The one remarkable thing about the 2535 apportionment is that Ellisium lost a seat while its population went up and Betta gained that seat while its population remained unchanged!



Example 4.6 is a rather trite story illustrating a fundamental paradox: Under Hamilton’s method, it is possible for a state with a positive population growth rate to lose one (or more) of its seats to another state with a smaller (or zero) population growth rate. Once again, Hamilton’s reliance on the residues to allocate the surplus seats is its undoing.

But wait, there is one more!

Hamilton’s Method-Population Paradox


In 1907 Oklahoma joined the Union. Prior to Oklahoma becoming a state, there were 386 seats in the House of Representatives. At the time the fair apportionment to Oklahoma was five seats, so the size of the House of Representatives was changed from 386 to 391. The point of adding these five seats was to give Oklahoma its fair share of seats and leave the apportionments of the other states unchanged.

The New-States Paradox


However, when the new apportionments were calculated, another paradox surfaced: Maine’s apportionment went up from three to four seats and New York’s went down from 38 to 37 seats. The perplexing fact that the addition of a new state with its fair share of seats can, in and of itself, affect the apportionments of other states is called the new-states paradox.

The New-States Paradox


The Metro Garbage Company has a contract to provide garbage collection and recycling services in two districts of Metropolis, Northtown (with 10,450 homes) and the much larger Southtown (89,550 homes). The company runs 100 garbage trucks, which are apportioned under Hamilton’s method according to the number of homes in the district. Aquick calculation shows that the standard divisor is SD = 1000 homes, a nice, round number which makes the rest of the calculations (shown in Table 4-12) easy.

Example 4.7 Garbage Time


As a result of the apportionment, 10 garbage trucks are assigned to service Northtown and 90 garbage trucks to service Southtown.



Now imagine that the Metro Garbage Company is bidding to expand its territory by adding the districtof Newtown (5250 homes) to its service area. In its bid to the City Council the company promises to buy five additional garbage trucks for the Newtown run so that its service to the other two districts is not affected. But when the new calculations (shown in Table 4-13) are carried out, there is a surprise: One of the garbage trucks assigned to Southtown has to be reassigned to Northtown!



Notice that the standard divisor has gone up a little and is now approximately 1002.38.



There are two key lessons we should take from this section:

Hamilton’s Method

(1) In terms of fairness, Hamilton’s method leaves a lot to be desired; and

(2) the critical flaw in Hamilton’s method is the way it handles the surplus seats. Clearly, there must be a better apportionment method.

chapter 4: the mathematics of sharing 4.1 apportionment problems and apportionment methods

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